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a. \(\sqrt{\frac{y}{5x^3}}=\sqrt{\frac{5xy}{25x^4}}=\frac{\sqrt{5xy}}{25x^2}\)
b\(\sqrt{\frac{5}{x\left(1-\sqrt{2}\right)}}=\sqrt{\frac{5\times x\left(1+\sqrt{2}\right)}{x^2\left(1-\sqrt{2}\right)\left(1+\sqrt{2}\right)}}=\sqrt{\frac{-5\times x\left(1+\sqrt{2}\right)}{x^2}}=-\frac{\sqrt{-5\times x\left(1+\sqrt{2}\right)}}{x}\)
c.\(\sqrt{\frac{x-1}{2\left(\sqrt{x}-1\right)}}=\sqrt{\frac{\sqrt{x}+1}{2}}=\frac{\sqrt{2\sqrt{x}+2}}{2}\)
d.\(a\sqrt{\frac{4}{a}}=\sqrt{\frac{4a^2}{a}}=\sqrt{4a}=2\sqrt{a}\)
e.\(2\sqrt{\frac{1}{-a}}=2\sqrt{\frac{-a}{a^2}}=-\frac{2}{a}\sqrt{-a}\left(\text{ do a< 0}\right)\)\(2\sqrt{\frac{1}{-a}}=2\sqrt{\frac{-a}{a^2}}=-\frac{2}{a}\sqrt{-a}\)( do a <0)
f.\(\sqrt{\frac{2}{x-1}-\frac{1}{\left(x-1\right)^2}}=\sqrt{\frac{2\left(x-1\right)-1}{\left(x-1\right)^2}}=\frac{\sqrt{2x-3}}{\left|x-1\right|}\)
a)\(=-\sqrt{\left(\frac{a}{b}\right)^2\cdot\frac{b}{a}}\)
\(=-\sqrt{\frac{a^2}{b^2}\cdot\frac{b}{a}}\)
\(=-\sqrt{\frac{a}{b}}\)
EM thử thôi, ko chắc đâu ạ:( Sai thì xin thông cảm cho ạ.
1) \(\sqrt{\frac{2}{3-\sqrt{5}}}=\sqrt{\frac{2\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}}=\sqrt{\frac{6+2\sqrt{5}}{4}}=\frac{\sqrt{6+2\sqrt{5}}}{2}\)
2) \(\sqrt{\frac{a-4}{2\left(\sqrt{a}-2\right)}}=\sqrt{\frac{\left(a-4\right)\left(\sqrt{a}+2\right)}{2\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}}\)
\(=\sqrt{\frac{\left(a-4\right)\left(\sqrt{a}+2\right)}{2\left(a-4\right)}}\)
3) \(\sqrt{\frac{1}{a\left(1-\sqrt{3}\right)}}=\sqrt{\frac{1+\sqrt{3}}{a\left(1-\sqrt{3}\right)\left(1+\sqrt{3}\right)}}=\sqrt{\frac{1+\sqrt{3}}{a\left(1-3\right)}}=\sqrt{-\frac{1+\sqrt{3}}{2a}}\)
4) \(\sqrt{\frac{a}{4-2\sqrt{3}}}=\sqrt{\frac{a\left(4+2\sqrt{3}\right)}{\left(4-2\sqrt{3}\right)\left(4+2\sqrt{3}\right)}}=\sqrt{\frac{4a+2a\sqrt{3}}{16-12}}=\sqrt{\frac{4a+2a\sqrt{3}}{4}}=\frac{\sqrt{4a+2a\sqrt{3}}}{2}\)
a) \(\sqrt{\frac{3}{125}}=\frac{\sqrt{3.125}}{125}=\frac{\sqrt{375}}{125}=\frac{5\sqrt{15}}{125}=\frac{\sqrt{15}}{25}\)
b) \(\sqrt{\frac{3}{2a^3}}=\frac{\sqrt{3.2a^3}}{2a^3}=\frac{\sqrt{6a^3}}{2a^3}\)
c) \(\sqrt{\frac{\left(1-\sqrt{3}\right)^2}{27}}=\frac{\sqrt{27\left(1-\sqrt{3}\right)^2}}{27}=\frac{3.\left(\sqrt{3}-1\right)\sqrt{3}}{27}=\frac{\left(\sqrt{3}-1\right)\sqrt{3}}{9}\)
d) \(\sqrt{\frac{11}{540}}=\frac{\sqrt{11.540}}{540}=\frac{\sqrt{5940}}{50}=\frac{6\sqrt{165}}{50}=\frac{3\sqrt{165}}{25}\)
Khử mẫu biểu thức chứa căn ms đúng
\(\sqrt{\frac{\left(1+\sqrt{2}\right)^3}{27}}=\sqrt{\frac{\left(1+\sqrt{2}\right)^2\cdot\left(1+\sqrt{2}\right)}{3^2\cdot3}}=\frac{1+\sqrt{2}}{3}\cdot\sqrt{\frac{1+\sqrt{2}}{3}}\)
\(=\frac{1+\sqrt{2}}{3}\cdot\frac{\sqrt{3\cdot\left(1+\sqrt{2}\right)}}{3}=\frac{1+\sqrt{2}}{9}\cdot\sqrt{3+3\sqrt{2}}\)
Bài 2:
a: \(=\sqrt{\left(\dfrac{1}{5a}\right)^2}=\dfrac{1}{\left|5a\right|}=\dfrac{-1}{5a}\)
b: \(=\dfrac{1}{3}\cdot15\cdot\left|a\right|=5\left|a\right|\)
Lời giải:
\(\sqrt{\frac{(1+\sqrt{2})^3}{27}}=\sqrt{\frac{(1+\sqrt{2})^3}{3^3}}=\sqrt{\frac{3(1+\sqrt{2})^3}{3^4}}\)
\(=\frac{(1+\sqrt{2})\sqrt{3+3\sqrt{2}}}{9}\)
\(ab\sqrt{\frac{1}{a}+\frac{1}{b}}=\sqrt{(ab)^2(\frac{1}{a}+\frac{1}{b})}=\sqrt{ab^2+a^2b}\)